Difference between revisions of "Selftest: Cross product"

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{ '''The following vectors are given.'''
 
{ '''The following vectors are given.'''
  
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[[File:Vektorrechnung_Vektorprodukt.jpg|300px]]
+
[[File:Vectoralgebra_crossproduct.jpg‎|300px]]
  
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Bitte markieren Sie alle richtigen Aussagen mit Berücksichtigung der gegebenen Vektoren.
+
Please mark the correct statements sonsidering the the given vectors.
 
}
 
}
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+ Der Vektor <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> steht senkrecht auf der Fläche, die von den Vektoren <math>\vec{\mathbf{a}}</math> und <math>\vec{\mathbf{b}}</math> aufgespannt wird.
+
+ The vector <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> is perpendicular to the plane that is spanned by the vectors <math>\vec{\mathbf{a}}</math> and <math>\vec{\mathbf{b}}</math>.
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- Der Betrag des vektoriellen Produkts ist gleich der Fläche des Parallelogramms, das von <math> \vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> aufgespannt wird.
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- The magnitude of the cross product equals the area of the parallelogram that is spanned by the vectors <math>\vec{\mathbf{a}}</math> and <math>\vec{\mathbf{b}}</math>.
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||Da die Formel des Vektorprodukts nicht nur das vektorielle Produkt, sondern auch den Sinus des eingeschlossenen Winkels enthält, ist diese Antwort falsch. Weitere Erklärung siehe [[Rechte Hand Regel 1]]
+
||The equation of the magnitude of the cross product not only includes the dot product but also the sine of the angle between the vectors. So this answer is wrong.
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+ Die Vektoren <math>\vec{\mathbf{a}}</math>, <math>\vec{\mathbf{b}}</math> und <math> \vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> bilden ein Rechtssystem, das heißt, sie sind angeordnet wie Daumen, Mittelfinger und Zeigefinger der rechten Hand.
+
+ The vectors <math>\vec{\mathbf{a}}</math>, <math>\vec{\mathbf{b}}</math> and <math> \vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> form a rectangular coordinate system. So they are arranged as thumb, middle finger and forefinger of the right hand.
 
</quiz>
 
</quiz>
  
 
[[Category:Selftest]]
 
[[Category:Selftest]]
 
[[Category:Vectors]]
 
[[Category:Vectors]]

Revision as of 17:07, 23 May 2014

← Previous exercise: Dot product Exercises for chapter {{{chapter}}} | Article: Vector algebra Next exercise: Introduction to vector algebra →
Point added for a correct answer:  
Points for a wrong answer:
Ignore the questions' coefficients:

1. Please mark the right transforms:

\vec{\mathbf{a}}\times\vec{\mathbf{b}}= 0 \text{ for } \vec{\mathbf{a}} \upuparrows\vec{\mathbf{b}}
\vec{\mathbf{a}}\times\vec{\mathbf{b}}= 0 \text{ for } \vec{\mathbf{a}}\bot\vec{\mathbf{b}}
→The magnitude of the cross product results from the following equation: \vec{\mathbf{a}}\times\vec{\mathbf{b}} =\vec{\mathbf{c}}\text{ with}\left|\vec{\mathbf{c}}\right| = ab\sin\alpha. In this case the angle is 90° and so the magnitude is |ab|.
\vec{\mathbf{a}}\times\vec{\mathbf{b}}= \vec{\mathbf{e}}_cab \text{ for } \vec{\mathbf{a}}\uparrow\downarrow\vec{\mathbf{b}}
→The magnitude of the cross product results from the following equation: \vec{\mathbf{a}}\times\vec{\mathbf{b}}=\vec{\mathbf{c}}\text{ with}\left|\vec{\mathbf{c}}\right|= ab\sin\alpha. Here the angle is 180° and the related sine is zero. So the result is 0.

2. Please mark the right transforms: (x,y,z are coordinate axes, a and b are arbitrary

\vec{\mathbf{b}} \times \vec{\mathbf{a}}=-(\vec{\mathbf{a}}\times \vec{\mathbf{b}})
\vec{\mathbf{y}}\times\vec{\mathbf{x}}=-\vec{\mathbf{z}}
The cross product is commutative.
→The first answer implies that the cross product is not commutative. This can easily be proved by the right-hand-rule. Further information: see Cross product

3. Please solve the following exercise:

\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}=\vec{\mathbf{e}}_x+\vec{\mathbf{e}}_y+\vec{\mathbf{e}}_z
→ To compute the cross product the component representation \vec{\mathbf{a}} \times \vec{\mathbf{b}} =(a_2 b_3 - a_3 b_2) \vec{\mathbf{e}}_1 + (a_3 b_1 - a_1 b_3) \vec{\mathbf{e}}_2 + (a_1 b_2 - a_2 b_1) \vec{\mathbf{e}}_3 is used.

4. Please solve the following exercise:

\begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix}=\vec{\mathbf{e}}_x+\vec{\mathbf{e}}_y+\vec{\mathbf{e}}_z
→ To compute the cross product the component representation \vec{\mathbf{a}} \times \vec{\mathbf{b}} =(a_2 b_3 - a_3 b_2) \vec{\mathbf{e}}_1 + (a_3 b_1 - a_1 b_3) \vec{\mathbf{e}}_2 + (a_1 b_2 - a_2 b_1) \vec{\mathbf{e}}_3 is used.

5. Please solve the following exercise:

\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\vec{\mathbf{e}}_x+\vec{\mathbf{e}}_y+\vec{\mathbf{e}}_z
→ To compute the cross product the component representation \vec{\mathbf{a}} \times \vec{\mathbf{b}} =(a_2 b_3 - a_3 b_2) \vec{\mathbf{e}}_1 + (a_3 b_1 - a_1 b_3) \vec{\mathbf{e}}_2 + (a_1 b_2 - a_2 b_1) \vec{\mathbf{e}}_3 is used.

6. The following vectors are given.

Vectoralgebra crossproduct.jpg

Please mark the correct statements sonsidering the the given vectors.

The vector \vec{\mathbf{a}}\times\vec{\mathbf{b}} is perpendicular to the plane that is spanned by the vectors \vec{\mathbf{a}} and \vec{\mathbf{b}}.
The magnitude of the cross product equals the area of the parallelogram that is spanned by the vectors \vec{\mathbf{a}} and \vec{\mathbf{b}}.
→The equation of the magnitude of the cross product not only includes the dot product but also the sine of the angle between the vectors. So this answer is wrong.
The vectors \vec{\mathbf{a}}, \vec{\mathbf{b}} and \vec{\mathbf{a}}\times\vec{\mathbf{b}} form a rectangular coordinate system. So they are arranged as thumb, middle finger and forefinger of the right hand.

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